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3 years ago

Why do the plants at the bottom of a pond grow better than plant on the bottom of a lake

Physics

1 answer:

yuradex [85]3 years ago

8 0

Because pond water is the most freshest water than lake water

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How many centimeters are there in meter? b. 10 c. d 1000 e. 10000 100 2. A centimeter is equal to 1 inch b.½inch C 1/2.54 inch

astra-53 [7]

Answer:

a) There are 100 centimeters in 1 meter.

b) $\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

Explanation:

a) We have the conversion

1 m = 100 cm

So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

$1cm=\frac{1}{2.54}inch$

$\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

8 0

2 years ago

2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o

Nadya [2.5K]

B is appropriate answer

7 0

3 years ago

Which of the following statements is true? there are only about 100 different kinds of atoms that combine to form all substances

TEA [102]

THE ANSWER IS : THERE ARE ONLY ABOUT 100 DIFFERENT KINDS OF ATOMS THAT COMBINE TO FORM ALL SUBSTANCES

Explanation:

3 0

2 years ago

The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and 780 mbars at the end. Neglecting the eff

lara31 [8.8K]

We could use the change of pressure to calculate for the height climbed by the mountain hiker. The change of pressure is given by

p = rho * g * h, where p is the change of pressure, rho is the air density, g is the acceleration due to gravity, and h is the height.

Using the conversion 1 mbar = 100 Pa,

(930 - 780)(100) = (1.20)(9.80)h
15000 = 1.20*9.80*h

h = 1.28 km

6 0

3 years ago

An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro

kupik [55]

Answer:

a) $Q_{in} = 13.742\,kW$ , b) $\Delta S = 370.15\,\frac{kJ}{K}$

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

$Q_{in} +U_{sys,1} - U_{sys,2} = 0$

$Q_{in} = U_{sys,2} - U_{sys,1}$

$Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})$

$Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)$

$Q_{in} = 13.742\,kW$

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

$\Delta S = \frac{Q_{in}}{T_{in}}$

$\Delta S = \frac{13.742\,kJ}{370.15\,K}$

$\Delta S = 370.15\,\frac{kJ}{K}$

3 0

3 years ago